How to prove that $R/I \otimes_R M \cong M / IM$ 
Possible Duplicate:
Showing that if $R$ is local and $M$ an $R$-module, then $M \otimes_R (R/\mathfrak m) \cong M / \mathfrak m M$. 

In one of the answers to one of my previous questions the following claim was mentioned:
$R/I \otimes_R M \cong M / IM$
So I tried to prove it. Can you help me finish my proof? Thanks!

We recall that $M \otimes_R -$ is a covariant right-exact functor that it is exact if $M$ is flat and we observe that the following is an exact sequence:
$$ 0 \to IM \xrightarrow{i} M \xrightarrow{\pi} M / IM \to 0$$
$R/I$ is an $R$ module since $R/I$ is a subring of $R$ closed with respect to multiplication from $R$ but it is not necessarily flat hence we only get exactness on one side:
$$  (R/I) \otimes_R IM \xrightarrow{id \otimes i} (R/I) \otimes_R M \xrightarrow{id \otimes \pi} (R/I) \otimes_R M / IM \to 0$$
Then $$ \mathrm{Im(id \otimes \pi)} = (R/I) \otimes_R M / IM  \cong (R/I) \otimes_R M / \mathrm{Ker} (id \otimes i) $$
so we want to show two things: 
(i) that $\mathrm{Ker} (id \otimes i) = \{0\}$. To this end let $r + I \otimes im \in (R/I) \otimes_R  IM $ and assume $ id \otimes i(r + I \otimes im ) = r + I \otimes im = 0 + I \otimes 0$. I'm not sure how to proceed from here. 
(ii) and that $M /IM \cong (R/I) \otimes_R M / IM $
Can you help me? Thanks. 
 A: In general, playing around with elements of a tensor product is something that should be avoided if you can help it (defining maps from a tensor product is also usually better done by defining them from the product and then checking that they are in fact $R$-balanced and the using the universal property). A proof which avoids dealing with explicit elements of the tensor product can be done in the following way:
Consider the map $R/I\times M\to M/IM$ given by $(r+I,m)\mapsto rm+IM$. This is well-defined, as the image of any two representatives of $r+I$ differ by an element of $IM$. Since it is $R$-balanced, it descends to a map from the tensor product. Now consider the map $M \to R/I\otimes M$ given by sending $m\mapsto (1+I)\otimes m$. This descends to the quotient $M/IM$ by the 1st isomorphism theorem, and then it is easy to see that these maps are mutual inverses.
A: Consider 
$$
  0 \to I \to R \to R/I \to 0
$$
Apply the right exact functor $-\otimes_R M$, and you get 
$$
   I\otimes_R M \to R\otimes_R M \to (R/I)\otimes_R M \to 0
$$
But $R\otimes_R M$ is canonically identified with $M$ by $a\otimes m \mapsto am$. Then $I\otimes_R M \to R\otimes_R M = M$ is $a\otimes m \mapsto am$ so, by definition, its image is $IM$. So the exactness of the sequence tells you that
$$
  M/IM \cong (R/I)\otimes_R M 
$$
